Proving that $x\mapsto \sum_{y\in A\cap\left(0,x\right]}r\left(y\right)$ is right-continuous Let $r:A\rightarrow\left(0,\infty\right)$ be defined on a countable
infinite subset $A\subseteq\mathbb{R}$. Let $T:\left(0,\infty\right)\rightarrow\mathbb{R}, \ T(x)=\sum_{y\in A\cap\left(0,x\right]}r\left(y\right)<\infty$ and assume $T(x)<\infty$ for all $x$. I want to show that $T$
is right-continuous at every point. This may seem intuitive, but
I couldn't manage to prove it rigorously: Take an arbitrary point
$t>0$ and an arbitrary sequence $\left(t_{n}\right)_{n}$ converging
from the right to it. I have to show then, that the sequence of sums
$\left(\sum_{y\in A\cap\left(0,t_{n}\right]}r\left(y\right)\right)_{n}$
converges to the sum $\sum_{y\in A\cap\left(0,t\right]}r\left(y\right)$. 
I managed to prove (by assuming that it doesn't converge) that there
exists is a subsequence of the sequence of sums $\left(\sum_{y\in A\cap\left(t,t_{n}\right]}r\left(y\right)\right)_{n}$
that is bounded from below by some $\varepsilon>0$ and is strictly decreasing,
meaning it converges to the $\inf$ of all the sums in the subsequence.
But that doesn't yet give me a contradiction (although I have the feeling that I'm very near)...Can you please tell
me how to get to a contradiction from here ?
 A: Fix $x,\epsilon>0$. Then $T(x+1) = \sum_{y \in A \cap (0,x+1]} r(y) < \infty$. Enumerating $A \cap (x,x+1] =  \{a_i\}$ this tells us $\sum a_i < \infty$ so $\sum_{i=n}^\infty a_i \to 0$ as $n \to \infty$. Choose $N_\epsilon$ large enough that $\sum_{i=N_\epsilon}^\infty a_i < \epsilon$. Then choose $\delta = \min\{\min_{i\leq N_\epsilon} |x-a_i|,1\}>0$. Then for any $y \in (x,x+\delta)$ we have
$$
T(y)-T(x) = \sum_{z \in A \cap (x,y]} r(z) \leq \sum_{z \in A \cap (x,\delta]} r(z) \leq \sum_{i=N_\epsilon}^\infty a_i < \epsilon.
$$
and it follows that $T$ is right continuous.
A: This is the series tail rule hiding in a different form. Normally, you have something like
$$ \sum_{n=1}^{\infty} a_n < \infty \implies \forall\epsilon>0\ \exists k\ \sum_{n=k}^{\infty} a_n < \epsilon$$
But in your case, the infinite input values are crammed up in some set $A$ in a way that might not be spaced out nicely. Well, no matter - all the series tail rule says is just that you can drop a finite number of summands to get an arbitrarily small sum. In particular, for any $\epsilon > 0$ there exists some $n$ such that
$$ \sum_{y \in A \cap \left(t, t_n\right]} r(y) < \epsilon $$
(that is, $t_n$ is close enough to $t$ that we dropped the finite number of values necessary to reach this $\epsilon$, as they were all to the right of $t_n$), giving us the desired result.
